Here's an implementation
of the program in C:
```c
#include <stdio
.h>
#include <stdlib
.h>
#include <pthread
.h>
#include <math
.h>
// Global variables for storing statistical values
double average = 0
.0;
int
minimum = 0;
int maximum = 0;
double median = 0
.0;
double std_dev = 0
.0;
// Struct for passing arguments to worker threads
typedef struct {
int* numbers;
int count;
} thread_args;
// Worker thread for calculating average
void* calculate_average(void* args) {
thread_args* targs = (thread_args*) args;
int* numbers = targs
->numbers;
int count = targs
->count;
double
sum = 0
.0;
for (int i = 0; i < count; i++) {
sum += numbers[i];
}
average =
sum / count;
pthread_exit(NULL);
}
// Worker thread for calculating
minimum
void* calculate_
minimum(void* args) {
thread_args* targs = (thread_args*) args;
int* numbers = targs
->numbers;
int count = targs
->count;
minimum = numbers[0];
for (int i = 1; i < count; i++) {
if (numbers[i] <
minimum) {
minimum = numbers[i];
}
}
pthread_exit(NULL);
}
// Worker thread for calculating maximum
void* calculate_maximum(void* args) {
thread_args* targs = (thread_args*) args;
int* numbers = targs
->numbers;
int count = targs
->count;
maximum = numbers[0];
for (int i = 1; i < count; i++) {
if (numbers[i] > maximum) {
maximum = numbers[i];
}
}
pthread_exit(NULL);
}
// Worker thread for calculating median
void* calculate_median(void* args) {
thread_args* targs = (thread_args*) args;
int* numbers = targs
->numbers;
int count = targs
->count;
int* sorted_numbers = malloc(count * size
of(int));
for (int i = 0; i < count; i++) {
sorted_numbers[i] = numbers[i];
}
for (int i = 0; i < count
- 1; i++) {
for (int j = i + 1; j < count; j++) {
if (sorted_numbers[i] > sorted_numbers[j]) {
int temp = sorted_numbers[i];
sorted_numbers[i] = sorted_numbers[j];
sorted_numbers[j] = temp;
}
}
}
if (count % 2 == 0) {
median = (sorted_numbers[count / 2
- 1] + sorted_numbers[count / 2]) / 2
.0;
} else {
median = sorted_numbers[count / 2];
}
free(sorted_numbers);
pthread_exit(NULL);
}
// Worker thread for calculating standard deviation
void* calculate_std_dev(void* args) {
thread_args* targs = (thread_args*) args;
int* numbers = targs
->numbers;
int count = targs
->count;
double
sum = 0
.0;
for (int i = 0; i < count; i++) {
sum += pow(numbers[i]
- average, 2);
}
std_dev = sqrt(
sum / count);
pthread_exit(NULL);
}
int main(int argc, char* argv[]) {
// Parse command line arguments
if (argc < 2) {
printf("Usage: %s <number1> <number2>
... <numberN>\n", argv[0]);
return 1;
}
int count = argc
- 1;
int* numbers = malloc(count * size
of(int));
for (int i = 0; i < count; i++) {
numbers[i] = atoi(argv[i + 1]);
}
// Create worker threads
pthread_t threads[5];
thread_args targs = {numbers, count};
pthread_create(&threads[0], NULL, calculate_average, &targs);
pthread_create(&threads[1], NULL, calculate_
minimum, &targs);
pthread_create(&threads[2], NULL, calculate_maximum, &targs);
pthread_create(&threads[3], NULL, calculate_median, &targs);
pthread_create(&threads[4], NULL, calculate_std_dev, &targs);
// Wait for worker threads to complete
for (int i = 0; i < 5; i++) {
pthread_join(threads[i], NULL);
}
// Print results
printf("The average value is %
.2f\n", average);
printf("The
minimum value is %d\n",
minimum);
printf("The maximum value is %d\n", maximum);
printf("The median value is %
.2f\n", median);
printf("The standard deviation is %
.2f\n", std_dev);
// Clean up
free(numbers);
return 0;
}
```
The program takes a list
of numbers as command line arguments, creates five worker threads to calculate statistical values, and then outputs the results once the worker threads have completed their calculations
. Note that the median calculation uses a simple sorting algorithm to sort the list
of numbers first
. This may not be efficient for very large
lists of numbers, but it works well enough for small
lists. Also note that the standard deviation calculation uses the formula for sample standard deviation, which divides by N
-1 instead
of N to correct for bias
.
本文介绍了一个算法问题，即如何找到两个人列表中共同喜爱的餐厅，并且这些餐厅在两个列表中的索引之和最小。通过使用字典存储一个列表的数据，可以高效地解决此问题。
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